Quadratic Equations — SBI Clerk Prelims Study Notes
Overview
The format is highly standardized: you're given two quadratic equations (one in x, one in y), asked to solve both, and then compare the roots to determine the relationship between x and y.
This topic is a scoring opportunity because the question pattern is predictable and the solving technique is mechanical. Once you master the factorization method and the comparison rules, you can solve each question in under 60 seconds. The key challenge isn't mathematical difficulty—it's speed and accuracy under time pressure.
Students who invest in learning the shortcut method for factorization and memorize the five comparison outcomes reliably score 4–5 out of 5 on this section. Treat this as a "must-score" topic in your Prelims preparation.
Key Concepts
- Standard form: A quadratic equation is ax² + bx + c = 0, where a, b, c are constants and a ≠ 0. SBI Clerk questions typically have a = 1 (monic form) for easier solving.
- Two roots: Every quadratic equation yields exactly two roots (which may be equal, distinct, or complex). In bank exams, roots are almost always real integers or simple fractions.
- Sum and product of roots: For x² + bx + c = 0, the sum of roots = −b and the product of roots = c. This helps verify your factorization.
- Factorization method: Split the middle term into two parts whose product equals a×c and whose sum equals b. This is the fastest solving approach for exam questions.
- Comparison logic: After finding roots of both equations, compare all possible pairs. The relationship depends on whether one variable's roots are consistently greater, smaller, or overlap with the other's.
- Five possible answers: The standard options are x > y, x < y, x ≥ y, x ≤ y, or "cannot be determined" (when ranges overlap without a clear relationship).
Formulas / Key Facts
For equation ax² + bx + c = 0:
Sum of roots = −b/a
Product of roots = c/a
Discriminant = b² − 4ac (positive means real distinct roots, zero means equal roots)
Factorization shortcut: Find two numbers whose product = a×c and sum = b, then split the middle term.
Comparison rules (after finding roots x₁, x₂ and y₁, y₂):
- If both x values > both y values → x > y
- If both x values < both y values → x < y
- If both x values ≥ both y values (with at least one equality) → x ≥ y
- If both x values ≤ both y values (with at least one equality) → x ≤ y
- If ranges overlap → Cannot be determined
Sign pattern for factorization:
- If c is positive: both factors have the same sign (determined by sign of b)
- If c is negative: factors have opposite signs
Worked Examples
Example 1: Equation I: x² − 7x + 12 = 0 Equation II: y² − 9y + 20 = 0
Solution: For x² − 7x + 12 = 0: Find two numbers with product = 12 and sum = 7 → 3 and 4 So x² − 3x − 4x + 12 = 0 → x(x−3) − 4(x−3) = 0 → (x−3)(x−4) = 0 x = 3 or x = 4
For y² − 9y + 20 = 0: Find two numbers with product = 20 and sum = 9 → 4 and 5 y = 4 or y = 5
Comparison: x values are {3, 4}, y values are {4, 5} x = 3 < both y values; x = 4 equals one y value Since x values ≤ y values → x ≤ y
Example 2: Equation I: x² + 5x + 6 = 0 Equation II: y² + 7y + 10 = 0
Solution: For x² + 5x + 6 = 0: Product = 6, sum = 5 → 2 and 3 (x+2)(x+3) = 0 → x = −2 or x = −3
For y² + 7y + 10 = 0: Product = 10, sum = 7 → 2 and 5 (y+2)(y+5) = 0 → y = −2 or y = −5
Comparison: x values are {−2, −3}, y values are {−2, −5} x = −2 equals y = −2; x = −3 > y = −5 but x = −3 < y = −2 Values overlap without consistent relationship → Cannot be determined
Example 3: Equation I: 2x² − 11x + 15 = 0 Equation II: 2y² − 13y + 21 = 0
Solution: For 2x² − 11x + 15 = 0: Product = 2×15 = 30, sum = 11 → 5 and 6 2x² − 5x − 6x + 15 = 0 → x(2x−5) − 3(2x−5) = 0 → (2x−5)(x−3) = 0 x = 5/2 = 2.5 or x = 3
For 2y² − 13y + 21 = 0: Product = 2×21 = 42, sum = 13 → 6 and 7 2y² − 6y − 7y + 21 = 0 → (2y−7)(y−3) = 0 y = 7/2 = 3.5 or y = 3
Comparison: x values are {2.5, 3}, y values are {3, 3.5} x = 2.5 < both y values; x = 3 equals y = 3 Since all x ≤ all y → x ≤ y
Common Mistakes
- Sign errors with negative coefficients: When b is positive in x² + bx + c = 0, roots are negative (since sum = −b). Students often forget the negative sign when writing final roots. → Always check: plug roots back to verify sum and product.
- Comparing only one pair of roots: Students find x = 3, y = 4 and conclude x < y, ignoring that x might have another value that's greater than y. → Always compare all four possible pairs (x₁ vs y₁, x₁ vs y₂, x₂ vs y₁, x₂ vs y₂).
- Confusing ≥ with > on boundary cases: When one root of x equals one root of y, but other comparisons are strict, the answer is ≥ or ≤, not > or <. → If there's any equality, use ≥ or ≤; pure inequalities only when no overlap exists.
- Wrong factor split for non-monic equations: For ax² + bx + c where a ≠ 1, students forget to multiply a×c for the product. → In 2x² − 11x + 15, find factors of 30 (not 15) that sum to 11.
- Rushing the "cannot be determined" option: Some students mark this when ranges partially overlap, but if one variable's maximum is still ≤ the other's minimum, a clear relationship exists. → Draw a number line mentally to visualize the ranges.
Quick Reference
- Split middle term: find two numbers with product = a×c and sum = b
- Negative b and positive c → both roots positive
- Positive b and positive c → both roots negative
- Negative c → roots have opposite signs
- Compare all four pairs; consistent inequality → definite answer; overlap → cannot determine
- Equal roots in comparison → use ≥ or ≤, not strict inequality